Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}x-y &= 3 \\ x+3y &= 2\end{align*}$
Explanation: Begin by moving the $x$ -term in the second equation to the right side of the equation. $3y = -x+2$ Divide both sides by $3$ to isolate $y$ $y = {-\dfrac{1}{3}x + \dfrac{2}{3}}$ Substitute this expression for $y$ in the first equation. $x-({-\dfrac{1}{3}x + \dfrac{2}{3}}) = 3$ $x + \dfrac{1}{3}x - \dfrac{2}{3} = 3$ Simplify by combining terms, then solve for $x$ $\dfrac{4}{3}x - \dfrac{2}{3} = 3$ $\dfrac{4}{3}x = \dfrac{11}{3}$ $x = \dfrac{11}{4}$ Substitute $\dfrac{11}{4}$ for $x$ back into the top equation. $ \dfrac{11}{4}-y = 3$ $\dfrac{11}{4}-y = 3$ $-y = \dfrac{1}{4}$ The solution is $\enspace x = \dfrac{11}{4}, \enspace y = -\dfrac{1}{4}$.